# Suppose that $\lim_ {l \rightarrow \infty} g(l)= \infty$

Let $q: \mathbb{R}→ \mathbb{R}$ be a continuous function. Suppose that $\lim_ {l \rightarrow \infty} g(l)= \infty$ and $\lim_{l \rightarrow - \infty} g(l)= - \infty$ Show that there exist $x$ and $y$ such that $g(x) \leq 0$ and $g(y) \geq 0$. Deduce that $g(l) =0$ has at least one solution.

This is what I was thinking...

Let $\lim_ {l \rightarrow \infty} g(l)= \infty$ and $\lim_{l \rightarrow - \infty} g(l)= - \infty$ be given. If $g(x) \geq 0$, $\forall x \in \mathbb{R}$ then $\lim_{x \rightarrow - \infty} g(x) \geq 0$. Hence, $- \infty \geq 0$, but this is a contradiction. Thus, there exists at least one $x \in \mathbb{R}$ such that $g(x) < 0$. Similarly, there exists at least one $y \in \mathbb{R}$ such that $g(y) > 0)$. Now, without loss of generality, assume that $x < y$. Since $g$ is continuous on $[x, y]$ and $g(x)< 0 < g(y)$ then by the intermediate value theorem there exists $c \in (x,y)$ such that $g(c)=0$. Hence, $g(x) = 0$ has at least one solution.

• You need to show some effort. – copper.hat May 21 '17 at 5:28
• Is this a correct proof now? – user6259845 May 22 '17 at 4:14
• Looks good to me. – copper.hat May 22 '17 at 4:18

Let $M_1 = 10$, then there is $N_1 > 0$ such that $x \ge N_1 \implies f(x) > M_1 = 10$, and let $M_2 = -10$, then there is $N_2 < 0$ such that if $x \le N_2 \implies f(x) < M_2 = -10$. We have $f(N_1) > 0, f(N_2) < 0$, and on the interval $[N_2,N_1]$, $f$ is continuous, thus by IVT, there is some $x \in$ this interval such that $f(x) = 0$.